What Is the Maximum Radius for a Loop So a Ball Stays on Track?

[pb23me]

Homework Statement

If the initial speed of the ball just before it enters the loop is 4 m/s, what is the largest value that the radius of the loop can have if the ball is to remain in contact with the track? Assume friction is negligible.I am supposed to figure this out with work and kinetic energy equations.

Homework Equations

Wnet= 1/2mv²_f-1/2mv²_i
Wg= mgy_f-mgy_i

The Attempt at a Solution

I can't really figure this out with those equations. I feel like i need more info.

 

[PhanthomJay]

You need to know about centripetal force and Newton's laws. If the radius gets too big, the contact (normal) force at the top of the loop will be 0, and the ball will fall off the track. Use this knowledge of centripetal force to find the maximum R with the use of the Conservation of mechanical energy equation.

 

[pb23me]

ok I've used the formula v=(rg)^1/2 and came up with 1.6m. I tried to set the equations equal to each other and solve for y_f but it seems like i need the final velocity at the top of the circle, but i don't know how to get that without the height...

 

[PhanthomJay]

ok I've used the formula v=(rg)^1/2 and came up with 1.6m.

that is the correct formula at the top of the circle when the Normal force is zero. I am not sure where or how you obtained that equation. But v is not 4m/s at the top of the circle, so the solution is incorrect.

I tried to set the equations equal to each other and solve for y_f but it seems like i need the final velocity at the top of the circle, but i don't know how to get that without the height...

Now use conservation of energy, using the initial mechanical energy at the bottom, and the final mechanical energy at the top. You will now have 2 equations with 2 unknowns (r and v_top), which you can solve by the method of your choice.

 

[pb23me]

I used the conservation of energy equation plugged in the value 2V_f/g for Y_f and solved for Vf getting 1.8 m/s. Then i went back and plugged that in and got .653m for r.

 

[PhanthomJay]

I used the conservation of energy equation plugged in the value 2V_f/g for Y_f

where did this come from? The equation and units are not correct.

and solved for Vf getting 1.8 m/s. Then i went back and plugged that in and got .653m for r.

retry using the conservation of energy equation again. Note that there is a speed at both the top and bottom of the loop. Note also that the velocity at the top is sq rt of Rg, and y_f at the top is??,

 

[pb23me]

I did it again an got .65m...r is not the same as Yf r is 1/2 Yf therefore it must be multiplied by two... I don't see what I'm doing wrong..

 

[PhanthomJay]

I did it again an got .65m...r is not the same as Yf r is 1/2 Yf therefore it must be multiplied by two... I don't see what I'm doing wrong..

You have already established that v_f^2 = rg (at the top of the loop) and V_i = 4 m/s (given, at the bottom of the loop). Now please write out the conservation of energy equation:

KE_i + PE_i = KE_f + PE_f

If your browser reads LaTex, that's

$$KE_i + PE_i = KE_f + PE_f$$

Now plug in the known and unknown variables for each term, noting as you said that y_f = 2r, and solve for r. Please write out each term so I can see what your error might be.

 

[pb23me]

y_f/2=r so 1/2v_i²=gy_f+1/2v_f² then 8=(5/4)*9.8*y_f
y_f=.65m
r=.33m

 

[PhanthomJay]

I used the conservation of energy equation plugged in the value 2V_f/g for Y_f and solved for Vf getting 1.8 m/s. Then i went back and plugged that in and got .653m for r.

This is wrong

I did it again an got .65m...r is not the same as Yf r is 1/2 Yf therefore it must be multiplied by two... I don't see what I'm doing wrong..

0.65 m is the value for what?

y_f/2=r so 1/2v_i²=gy_f+1/2v_f² then 8=(5/4)*9.8*y_f
y_f=.65m
r=.33m

This is right!